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3 years ago

9

A park in a subdivision has a triangular shape. Two adjacent sides of the park are 533 feet and 525 feet. The angle between the

sides is 53°. Find the area of the park to the nearest square foot

1 answer:

Anit [1.1K]3 years ago

8 0

Hopes this helps u good luck

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Which of the following is the graph of the solution<br> of 3x+9≤30 ?<br>

Alex777 [14]

Answer:

3x+9≤30 is x≤7

Step-by-step explanation:

4 0

4 years ago

Boats from all along the Atlantic coast were moored at a busy marina one summer day. Of the 39 boats at the marina, 13 were from

DIA [1.3K]

Answer:

The "experimental" probability is gotten in this "experiment" so far by looking at (# successes)/(# trials).

#S = 6 #T=15.

Step-by-step explanation:

Can you pleasehelp me?

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3 years ago

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago

Y-12=4y solve for y and show work plz

Katena32 [7]

First, you want to start by subtracting y from each side to get it isolated. After that, your equation should be
3y= -12
Then, divide each side by 3 to isolate the y. You should end up with y= -4

7 0

3 years ago

Read 2 more answers

Ella swims four times a week at the pool she swims the same number of laps on Monday, Wednesday , and Friday ,, and 15 laps on S

Nikitich [7]

Answer:

12

Step-by-step explanation:

51-15=

36

36/3=12

Have a great day!

3 0

3 years ago